Answer:
b+6
Problem:
If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
Step-by-step explanation:
We are given (b+c)/2=8 and d=3b-4.
We are asked to find (c+d)/2 in terms of variable, b.
We need to first solve (b+c)/2=8 for c.
Multiply both sides by 2: b+c=16.
Subtract b on both sides: c=16-b
Now let's plug in c=16-b and d=3b-4 into (c+d)/2:
([16-b]+[3b-4])/2
Combine like terms:
(12+2b)/2
Divide top and bottom by 2:
(6+1b)/1
Multiplicative identity property applied:
(6+b)/1
Anything divided by 1 is that anything:
(6+b)
6+b
b+6
Answer:
7.(1,0) and (-4,0)
8. (-3,0) and (-2,0)
Step-by-step explanation:
7.
When the function is equal to zero, we are looking for the zeros, or the x intercepts
Where it crosses the x axis are 1 and -4
(1,0) and (-4,0)
8. When the function is equal to zero, we are looking for the zeros, or the x intercepts
Where it crosses the x axis are -3 and -2
(-3,0) and (-2,0)
Answer:
B .`-(1)/(8) stackrel(+)(-) (isqrt(79))/(8)`
Answer : C. 512
step-by-step explanation :
- 8
- 8(square) = 8² = 8 × 8= 64
- 8(cube) = 8³ = 8 × 8 × 8 = 64 × 8
Answer:
c
Step-by-step explanation:
